Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{3z - 27}{z + 6} \times \dfrac{5z^2 + 20z - 105}{z^2 - 2z - 63} $
Solution: First factor out any common factors. $q = \dfrac{3(z - 9)}{z + 6} \times \dfrac{5(z^2 + 4z - 21)}{z^2 - 2z - 63} $ Then factor the quadratic expressions. $q = \dfrac {3(z - 9)} {z + 6} \times \dfrac {5(z + 7)(z - 3)} {(z + 7)(z - 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {3(z - 9) \times 5(z + 7)(z - 3) } {(z + 6) \times (z + 7)(z - 9) } $ $q = \dfrac {15(z + 7)(z - 3)(z - 9)} {(z + 7)(z - 9)(z + 6)} $ Notice that $(z + 7)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {15\cancel{(z + 7)}(z - 3)(z - 9)} {\cancel{(z + 7)}(z - 9)(z + 6)} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $q = \dfrac {15\cancel{(z + 7)}(z - 3)\cancel{(z - 9)}} {\cancel{(z + 7)}\cancel{(z - 9)}(z + 6)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac {15(z - 3)} {z + 6} $ $ q = \dfrac{15(z - 3)}{z + 6}; z \neq -7; z \neq 9 $